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Homework Help: Differantiation proof question exlanation problem

  1. Jan 21, 2009 #1
    if f(x) is differentiable on x_0
    prove that
    http://img102.imageshack.us/img102/3189/51290270sj1.th.gif [Broken]

    ??

    [itex] f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} [/itex]

    now let [itex] u = x+h \Rightarrow x = u-h [/itex]

    So [itex] f'(u-h) = \lim_{h \to 0} \frac{f(u-h+h)-f(u-h)}{h} = \lim_{h \to 0} \frac{f(u)-f(u-h)}{h} [/itex]

    Since [itex] h \to 0, f'(u-h) \to f'(u) [/itex]

    Therefore [itex] f'(x) = \lim_{h \to 0} \frac{f(x)-f(x-h)}{h} [/itex]

    [itex] f'(x)+f'(x) = 2f'(x) = \lim_{h \to 0}\left( \frac{f(x+h)-f(x)}{h} + \frac{f(x)-f(x-h)}{h} \right)[/itex]

    [itex] 2f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{h} [/itex]

    Hence, [itex] f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{2h} [/itex]


    i understand this solution but
    how to pick those variables??
    and why it didnt use the point x_0
    that they presented
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 21, 2009 #2
    there is another definition to limit that uses x_0 that i was given
    [itex]
    f'(x) = \lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}
    [/itex]

    is there a way to use this definition like you did before
    what variables to pick so it will go on the "h" defnition??
     
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